Question
\[N{{a}_{2}}{{B}_{4}}{{O}_{7}}\cdot 10{{H}_{2}}O\xrightarrow{\Delta }X\] \[\xrightarrow{\Delta }Y+Z\]
$$X, Y$$ and $$Z$$ in the reaction are
A.
$$X = N{a_2}{B_4}{O_7},Y = NaB{O_2},$$ $$Z = {B_2}{O_3}$$
B.
$$X = N{a_2}{B_4}{O_7},Y = {B_2}{O_3},$$ $$Z = {H_3}B{O_3}$$
C.
$$X = {B_2}{O_3},Y = NaB{O_2},$$ $$Z = B{\left( {OH} \right)_3}$$
D.
$$X = NaB{O_2},Y = {B_2}{O_3},$$ $$Z = B{\left( {OH} \right)_3}$$
Answer :
$$X = N{a_2}{B_4}{O_7},Y = NaB{O_2},$$ $$Z = {B_2}{O_3}$$
Solution :
\[N{{a}_{2}}{{B}_{4}}{{O}_{7}}\cdot 10{{H}_{2}}O\xrightarrow{\Delta }\underset{\left( X \right)}{\mathop{N{{a}_{2}}{{B}_{4}}{{O}_{7}}}}\,\] \[\xrightarrow{\Delta }\underset{\left( Y \right)}{\mathop{2NaB{{O}_{2}}}}\,+\underset{\left( Z \right)}{\mathop{{{B}_{2}}{{O}_{3}}}}\,\]